What is the de Brogile wavelength of an electon that strikes the back of the face of a TV screen at 1/10 the speed of light?
It is h/p where h is Planck's constant and p is the momentum of the electron.
I'm still not understand how to put it in the calculator, could you please explain?
h = 6.62606876*10^(-34) Js
mass = 9.10938185726*10^(-31) kg
speed = 0.1 c
c = 299792458 m/s
momentum is p = mass times speed
Thanks, Count Iblis!!!
To determine the de Broglie wavelength of an electron, we can use the de Broglie relation, which states that the wavelength (λ) of a particle is equal to Planck's constant (h) divided by the momentum (p) of the particle. Mathematically, it can be expressed as λ = h/p.
In this case, we need to find the momentum of the electron. Momentum (p) can be calculated as the product of the mass (m) and velocity (v) of the electron. Mathematically, p = m * v.
Given that the electron is traveling at 1/10 the speed of light, we can assume its velocity (v) to be 1/10 of the speed of light (c). The speed of light, denoted as 'c,' is approximately 3 x 10^8 meters per second (m/s).
So, v = (1/10) * c = (1/10) * (3 x 10^8 m/s) = 3 x 10^7 m/s.
The mass of an electron (m) is approximately 9.1 x 10^-31 kilograms (kg).
Now, we can calculate the momentum of the electron by multiplying its mass and velocity:
p = m * v = (9.1 x 10^-31 kg) * (3 x 10^7 m/s).
Once we have the momentum of the electron, we can calculate its de Broglie wavelength using the de Broglie relation:
λ = h/p,
where Planck's constant (h) is approximately 6.626 x 10^-34 joule-seconds (J·s).
Using this information, we can now find the de Broglie wavelength of the electron.